Question

Through the point $P(4,1)$ a line is drawn to meet the line $3 x-y=0$ at $Q$ where $P Q=\frac{11}{2 \sqrt{2}}$. Determine the equation of line.

• A. $x+y=5, x-7 y+3=0$
• B. $x-y=5, x-7 y+3=0$
• C. $x+y=5, x+7 y+3=0$
• D. $x-y=5, x+7 y+3=0$

Answer 1

Answer: (A)

By parametric form $Q\left(4+\frac{11}{2 \sqrt{2}} \cos \theta, 1+\frac{11}{2 \sqrt{2}} \sin \theta\right)$
it lies on $3 x-y=0 \Rightarrow 12+\frac{33}{2 \sqrt{2}} \cos \theta-1-\frac{11}{2 \sqrt{2}} \sin \theta=0$
$\Rightarrow 1+\frac{3}{2 \sqrt{2}} \cos \theta-\frac{\sin \theta}{2 \sqrt{2}}=0 \Rightarrow 3 \cos \theta-\sin \theta=-2 \sqrt{2}$
squaring both sides
$9 \cos ^2 \theta+\sin ^2 \theta-6 \sin \theta \cos \theta=8\left(\sin ^2 \theta+\cos ^2 \theta\right) \Rightarrow \cos ^2 \theta-6 \sin \theta \cos \theta-7 \sin ^2 \theta=0$
$7 \tan ^2 \theta+6 \tan \theta-1=0 \Rightarrow \tan \theta=-1, \frac{1}{7} .$
Hence required line are $x+y=5, x-7 y+3=0$